Question: The equation of a circle $C$ is $x^2+y^2-12x-6y+9 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-12x) + (y^2-6y) = -9$ $(x^2-12x+36) + (y^2-6y+9) = -9 + 36 + 9$ $(x-6)^{2} + (y-3)^{2} = 36 = 6^2$ Thus, $(h, k) = (6, 3)$ and $r = 6$.